By Cauchy's criterion of limit, show that $\lim_{x\to 0}(\sin{\frac{1}{x}}+x\cos{\frac{1}{x}})$ does not exist. By Cauchy's criterion of limit (not sequencial criterion), show that $$\lim_{x\to 0}(\sin{\frac{1}{x}}+x\cos{\frac{1}{x}})$$ does not exist.
Cauchy's criterion of limit 
$\lim_{x\to c}f(x)=l$ iff for every $\epsilon>0$, there exists $\delta$ such that $$|f(x_2)-f(x_1)|<\epsilon$$ for $0<|x_1-c|<\delta$ and $0<|x_2-c|<\delta$. 
Please suggest $x_2, x_1$ and help me to solve the problem.
 A: One part of Cauchy's Criterion says that

RESULT: If $\exists \epsilon>0$ such that $\forall \delta >0$, we can find $x_1,x_2$ satisfying $0<|x_1-a|<\delta$ and $0<|x_2-a|<\delta$ but $|f(x_2)-f(x_1)|\geq \epsilon$ then $\lim_{x\to a}f(x)$ does not exist. 

Let us write $$f(x)=\sin\frac{1}{x}+x\cos\frac{1}{x}.$$
Take $\epsilon\leq 2$. Let $\delta>0$. By Archimedean property,we can find $n\in\Bbb N$ such that $\frac{1}{n}<\delta$. Take
$$x_1=\frac{1}{\frac{3\pi}{2}+2\pi n}\qquad\text{and}\qquad x_2=\frac{1}{\frac{\pi}{2}+2\pi n}.$$ Notice that $0<x_1<\frac{1}{n}<\delta$ and $0<x_2<\frac{1}{n}<\delta$. Notice that $\sin x_2=1$, $\sin x_1=-1$, $\cos x_2=0$, and $\cos x_1=0$. With this, we get
$$|f(x_2)-f(x_1)|=|1-(-1)|=2\geq\epsilon.$$
Apply the result and we are done.
NOTE: That was the idea behind the hint given by @5xum. The only difference is that our delta and epsilon were interchanged. Since the OP want for clarification, I rather post an answer than to put all of these in the comments.
A: Hint:
You can find, for any $\epsilon > 0$, a value of $x_1$ such that $0<x_1<\epsilon$ and $\sin\frac1{x_1}=1$, and a value of $x_2$ such that $0<x_2<\epsilon$ and $\sin\frac{1}{x_2}=-1$. 
A: Note first that $$\lim_{x \to 0}x \cos \frac 1x=0$$ so you can safely focus on $\sin \frac 1x$. Now consider $x_n=\frac 1{2n\pi+\frac{\pi}2}$ and $y_n=\frac1{n\pi}$ for $n\to \infty$ and see what happens.
